Divide the following complex numbers. $ \dfrac{6-18i}{4-2i}$
Explanation: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${4+2i}$ $ \dfrac{6-18i}{4-2i} = \dfrac{6-18i}{4-2i} \cdot \dfrac{{4+2i}}{{4+2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(6-18i) \cdot (4+2i)} {(4-2i) \cdot (4+2i)} = \dfrac{(6-18i) \cdot (4+2i)} {4^2 - (-2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(6-18i) \cdot (4+2i)} {(4)^2 - (-2i)^2} = $ $ \dfrac{(6-18i) \cdot (4+2i)} {16 + 4} = $ $ \dfrac{(6-18i) \cdot (4+2i)} {20} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({6-18i}) \cdot ({4+2i})} {20} = $ $ \dfrac{{6} \cdot {4} + {-18} \cdot {4 i} + {6} \cdot {2 i} + {-18} \cdot {2 i^2}} {20} $ Evaluate each product of two numbers. $ \dfrac{24 - 72i + 12i - 36 i^2} {20} $ Finally, simplify the fraction. $ \dfrac{24 - 72i + 12i + 36} {20} = \dfrac{60 - 60i} {20} = 3-3i $